Question

$\sum_{r = 1 }^{n - 1} ( \frac{1}{ \sqrt{ {4n}^{2} - {r}^{2} } } ) \: as \: n \: \to \: \infty$

Find the sum of above.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\sum_{r=1}^{n-1}\sf \frac{1}{ \sqrt{ {4n}^{2} - {r}^{2} } } \: \: as \: n \: \to \: \infty$

can be rewritten as

$\rm \:  =  \: \displaystyle\sum_{r=1}^{n-1}\sf \frac{1}{ \sqrt{ {r}^{2} \bigg[4 - \dfrac{ {r}^{2} }{ {n}^{2} } \bigg]} } \: \: \: as \: n \: \to \: \infty$

$\rm \:  =  \: \displaystyle \lim_{n \: \to \: \infty }\displaystyle\sum_{r=1}^{n-1}\sf \frac{1}{r} \frac{1}{ \sqrt{\bigg[4 - \dfrac{ {r}^{2} }{ {n}^{2} } \bigg]} } \: \:$

Now, using limit as sum of definite integrals,

$\rm :\longmapsto\:\dfrac{r}{n} \: changes \: to \: x$

$\rm :\longmapsto\:\dfrac{1}{n} \: changes \: to \: dx$

$\rm :\longmapsto\:\displaystyle \sf \lim_{n \to \infty} \sum \: changes \: to \: \int$

$\rm :\longmapsto\: \: lower \: limit \: a \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{1}{n} = 0$

$\rm :\longmapsto\: \: upper \: limit \: b \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{n - 1}{n} = \: 1$

So, using all these, we get

$\rm \:  =  \: \displaystyle\int_{0}^{1} \sf \: \frac{dx}{ \sqrt{4 - {x}^{2} } }$

$\rm \:  =  \: \displaystyle\int_{0}^{1} \sf \: \frac{dx}{ \sqrt{ {2}^{2} - {x}^{2} } }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c \: }}}$

So, using this identity, we get

$\rm \:  =  \: {sin}^{ - 1} \dfrac{x}{2} \bigg| _{0}^{1}$

$\rm \:  =  \: {sin}^{ - 1} \dfrac{1}{2} - {sin}^{ - 1}0$

$\rm \:  =  \: \dfrac{\pi}{6}$

Hence,

$\rm\implies \:\:\boxed{\tt{ \displaystyle\sum_{r=1}^{n-1}\sf \frac{1}{ \sqrt{ {4n}^{2} - {r}^{2} } } \: \: as \: n \: \to \: \infty = \dfrac{\pi}{6} \: }}$

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MORE TO KNOW

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\sum_{r=1}^{n-1}\sf \frac{1}{ \sqrt{ {4n}^{2} - {r}^{2} } } \: \: as \: n \: \to \: \infty$

can be rewritten as

$\rm \:  =  \: \displaystyle\sum_{r=1}^{n-1}\sf \frac{1}{ \sqrt{ {r}^{2} \bigg[4 - \dfrac{ {r}^{2} }{ {n}^{2} } \bigg]} } \: \: \: as \: n \: \to \: \infty$

$\rm \:  =  \: \displaystyle \lim_{n \: \to \: \infty }\displaystyle\sum_{r=1}^{n-1}\sf \frac{1}{r} \frac{1}{ \sqrt{\bigg[4 - \dfrac{ {r}^{2} }{ {n}^{2} } \bigg]} } \: \:$

Now, using limit as sum of definite integrals,

$\rm :\longmapsto\:\dfrac{r}{n} \: changes \: to \: x$

$\rm :\longmapsto\:\dfrac{1}{n} \: changes \: to \: dx$

$\rm :\longmapsto\:\displaystyle \sf \lim_{n \to \infty} \sum \: changes \: to \: \int$

$\rm :\longmapsto\: \: lower \: limit \: a \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{1}{n} = 0$

$\rm :\longmapsto\: \: upper \: limit \: b \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{n - 1}{n} = \: 1$

So, using all these, we get

$\rm \:  =  \: \displaystyle\int_{0}^{1} \sf \: \frac{dx}{ \sqrt{4 - {x}^{2} } }$

$\rm \:  =  \: \displaystyle\int_{0}^{1} \sf \: \frac{dx}{ \sqrt{ {2}^{2} - {x}^{2} } }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } = {sin}^{ - 1} \frac{x}{a} + c \: }}}$

So, using this identity, we get

$\rm \:  =  \: {sin}^{ - 1} \dfrac{x}{2} \bigg| _{0}^{1}$

$\rm \:  =  \: {sin}^{ - 1} \dfrac{1}{2} - {sin}^{ - 1}0$

$\rm \:  =  \: \dfrac{\pi}{6}$

Hence,

$\rm\implies \:\:\boxed{\tt{ \displaystyle\sum_{r=1}^{n-1}\sf \frac{1}{ \sqrt{ {4n}^{2} - {r}^{2} } } \: \: as \: n \: \to \: \infty = \dfrac{\pi}{6} \: }}$

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